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Laurent Series
Adrian Down
October 27, 2005
1 e
F (z)
1.1 Statement
Theorem. Assume
- D be a simply connected domain.
- f is differentiable on D
- f (z) 6 = 0 ∀z ∈ D
Then
Then ∃F : D → C st
eF^ ≡ f F is differentiable
Note. This theorem says that f has a differentiable logarithm.
1.2 Motivating example
Example.
f (z) = z D = C r { 0 }
Assume for the sake of contradiction that F exists. Then Log z is differen- tiable on D ⇒ (^1) z has an anti-derivative in D ⇒ the path integral around the circle of radius 1 is 0. This shows that the assumption that D is simply connected is important.
1.3 Proof
Proof. Fix z 0 ∈ D. Since D is path connected, for any z ∈ D, choose a piecewise smooth path γ(z, z 0 ) from z 0 to z in D. We want to integrate a function along this path to get F. We take an aside to determine what function to integrate.
eF^ = f ⇒ F ′eF^ = f ′
⇒ F ′^ =
f ′ f
Thus we define,
F (z) =
γ(z,z 0 )
f ′ f
Because f (z) 6 = 0 for any z ∈ D, f^
′ f is continuous, and because^ D^ is sim- ply connected, integrals of continuous functions in D are path-independent. From previous results in the course, F is differentiable, and
F ′(z) =
f ′(z) f (z)
∀z ∈ D
We have constructed F differentiable on D, and
F ′^ =
f ′ f
We want to show that
eF^ (z)^ = f (z)
To show equality, we use the standard technique of showing that another function is constant. Consider
g =
eF f
If D is simply connected, and 0 ∈/ D, then we can use the theorem just proved to show that ∃ a differentiable logarithm in D, i.e. a function F : D → C such that
eF^ (z)^ = z ∀z ∈ D
Thus the previous theorem gives a condition on the existence of a differen- tiable logarithm.
2 Laurent series
2.1 Definitions
2.1.1 Doubly-infinite series
A Laurent series is one of the form
∑^ ∞
n=−∞
an(z − z 0 )n
Before rigorously defining Laurent series, we first have to discuss doubly infinite series, specifically what it means for some Laurent series to converge.
Definition.
∑^ ∞
n=−∞
cn cn ∈ C converges ⇔
∑^ ∞
n=
cn converges and
∑^ ∞
k=
c−k converges
Example.
cn =
1 n ≥ 0 − 1 n < 0
We have then
∑^ N
n=−N
cn = 1
so perhaps we should define
∑^ ∞
n=−∞
cn = +
However,
∑^2 N
n=−N
cn = N + 1 −−−→ N →∞
This limit depends on how it is evaluated. Thus using a partial sum defini- tion is a poor choice for convergence of doubly-infinite series. Checking the positive and negative n terms of the series for convergence separately ensures such difficulties do not arise.
2.1.2 Laurent series
Definition (Laurent Series). A Laurent series is an infinite series
∑^ ∞
n=−∞
an(z − z 0 )n^ z 0 ∈ C and an ∈ C
2.1.3 Annulus
Let 0 ≤ R 1 < R 2 ≤ ∞
Definition (Annulus). An annulus is an open set of the form
A = { z ∈ C : R 1 < | z − z 0 | < R 2 }
Note. • Usually both R 1 and R 2 are finite. This gives a donut-shaped area.
- We also allow the punctured disk, i.e. R 1 = 0.
- Recall the difference between convergence and absolute convergence. Convergence means
∀ > 0 , ∃N, M < ∞ st
n ≥ N ⇒
∑^ ∞
n=
an(z − z 0 )
< M
Uniform convergence means that M does not depend on z.
2.2.3 Proof
Proof. Take z 0 = 0 for simplicity. We are given that f is differentiable for R 1 < |z| < R 2. Remember that R 1 could be 0, and R 2 could be ∞. The most basic tool we have for understanding differentiable functions are path integrals. Thus we try to prove our formulation for f (z) in terms of a power series by considering integrals. Choose
R 1 < ρ 1 < |z| < ρ 2 < R 2
Call the circles of radius ρi Cρi The claim is
f (z) =
2 πı
Cρ 2
f (w) w − z
dw −
2 πı
Cρ 1
f (w) w − z
dw
We would like to use Cauchy’s theorem. To do so, take one more circle γ centered at z of small enough radius such that
ρ 1 < |w| < ρ 2 ∀w ∈ range of γ
Apply Cauchy’s theorem to these three circles in the following way. We want to show that for any function g differentiable in A r { z }, ∫
Cρ 2
g −
Cρ 1
g −
γ
g = 0
Once we have this, we can apply it to
g(w) =
f (w) w − z
to get
1 2 πı
Cρ 2
f (w) w − z
dz −
2 πı
Cρ 1
f (w) w − z
dw =
2 πı
γ
f (w) w − z
dw
We have seen that
1 2 πı
γ
f (w) w − z
dw = f (z)
To check the identity, use the generalized Cauchy’s theorem. Then we need only talk about winding numbers. It suffices to show
∀z 1 ∈/ A r { z } w(Cρ 2 , z 1 ) − w(Cρ 1 , z 1 ) − w(γ, z 1 ) = 0
There are three types of points in the annulus. The winding numbers of all three are easily computed by eye.
- The points outside of all three circles are clearly 0.
- For points in the interior of both Cρ 1 , and Cρ 2 , both paths contribute 1 to the winding number. However, we subtract these contributions, so that the winding number is 1 − 1 = 0, as desired.
- For z, it is traversed by the outer circle and γ, so we get 1 − 0 − 1.
To continue the proof, we have to convert the path integrals to coefficients of a power series. Use a fact from the previous proof in which we constructed the Taylor series from path integrals, here stated as a lemma,
Lemma. Assume
- ρ > 0
- Cρ = circle centered at 0 with radius ρ
- h(z) defined for |z| < ρ
- h continuous
3 Singularities
3.1 Definitions
Definition (Isolated singularity). Assume
- S is an open set
- f is differentiable on S
- z 0 ∈/ S
∃δ > 0 st { z : 0 < | z − z 0 | < δ } ⊂ S
Then f has an isolated singularity at z 0.
Note. The domain of f ⊃ a punctured disk centered at z 0 , but not z 0
Definition (Removable singularities). An isolated singularity at z 0 is said to be removable if ∃g differentiable in some neighborhood of z 0 such that g(z) = f (z) at every point of that neighborhood except z 0.
Note. • By defining the function at the point of a removable singularity, the function can be made differentiable. Removable singularities can result from specifying a bad domain for a function.
- Not all singularities are isolated. Consider,
Log : Cπ → C
The singularity at z 0 = 0 is not isolated.
